$A$ large tank filled with water to a height $h$ is to be emptied through a small hole at the bottom. The ratio of the time taken for the level to fall from $h$ to $h/2$ and that taken for the level to fall from $h/2$ to $0$ is:

$A$ fully filled open water tank has two holes on either side of its walls. One is a square hole of side $x \,cm$ at a depth of $2 \,m$ from the top, and the other hole is an equilateral triangle of side $4 \,cm$ at a depth of $6 \,m$ from the top. If the rate of flow of water is the same from both the holes, then the value of $x$ is: (in $\,cm$)

$A$ cylindrical tank having a large diameter is filled with water to a height $H$. $A$ hole of cross-sectional area $5 \,cm^2$ in the tank allows water to drain out. If the water drains out at the rate of $2 \times 10^{-3} \,m^3 \,s^{-1}$, then the value of $H$ is (acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,cm$)

$A$ vessel of area of cross-section $A$ has liquid to a height $H$. There is a hole at the bottom of the vessel having area of cross-section $a$. The time taken to decrease the level from $H_1$ to $H_2$ will be:

$A$ cylindrical tank has a hole of area $2 \,cm^2$ at its bottom. If water is poured into the tank from a tube above it at the rate of $100 \,cm^3 \,s^{-1}$, then the maximum height up to which water can rise in the tank is (Acceleration due to gravity, $g=10 \,ms^{-2}$) (in $cm$)

$A$ large tank filled with water to a height $h$ is to be emptied through a small hole at the bottom. The ratio of times taken for the level of water to fall from $h$ to $h/2$ and from $h/2$ to $0$ is